Problem: $f(x, y) = (1, 2x\ln(xy))$ $\text{curl}(f) = $
Solution: The formula for curl in two dimensions is $\text{curl}(f) = \dfrac{\partial Q}{\partial x} - \dfrac{\partial P}{\partial y}$, where $P$ is the $x$ -component of $f$ and $Q$ is the $y$ -component. Let's differentiate! $\begin{aligned} \dfrac{\partial Q}{\partial x} &= \dfrac{\partial}{\partial x} \left[ 2x\ln(xy) \right] \\ \\ &= 2\ln(xy) + 2x \left( \dfrac{y}{xy} \right) \\ \\ &= 2\ln(xy) + 2 \\ \\ \dfrac{\partial P}{\partial y} &= \dfrac{\partial}{\partial y} \left[ 1 \right] \\ \\ &= 0 \end{aligned}$ Therefore: $\begin{aligned} \text{curl}(f) &= 2\ln(xy) + 2 \end{aligned}$